| Basic Thermodynamics ~ J. Pohl © | www.THERMOspokenhere.com (195-E145) |
The sketch depicts water flowing from the left to the right without friction (we assume) through a horizontal pipe. A fixed "origin" is notated. Left of the origin the pipe diameter is d. An expansion is located at X = 0. The pipe dimeter thereafter is D. Suppose at the instant, t = t*, a point in the flow at X = XL(t*) is marked and observed to have the velocity vL(t*).
Prove the velocity of the water at X = XR(t*) is equal to (AL/AR)vL(t*).
♦ The answer to this problem is obvious. We're interested in the method of solution; not the answer. Let the system be the water bounded on the left by a plane at XL(t*)) and on the right by a plane at XR(t*). Both planes are moving. The points were noted at a superscripted time, t*, (which is usually set at, t* = 0+). Since the points move with the water, the mass between them is constant (by physical constraint of the pipe) and the derivative of that the mass so bounded equals zero. ( The notation, t*, is used because an observation is not made at time, t, a variable. However, after a conclusion is made with regard to a t*, it can be said that since t* was arbitrary, the results apply for all t).
| (1) 1 |
The system mass equals the integral of density times area between the left and right boundaries.
| (2) 2 |
To continue we see in the integrand that area, A, is discontinuous at X = 0. So we break the integral into two pieces, reverse the limits on the second integral and identify the areas as, AL and AR. The rearranged integral is:
| (3) 3 |
The densities and areas are constant so we perform the integrations then differentiate.
| (4) 4 |
or
| (5) 5 |
Recognizing dX/dt as linear speed (magnitude of velocity), a result obtained is:
| (6) 6 |
And our solution is:
| (7) 7 |
This is what we expected. The purpose of this was to show that the mathematics works.
